2 years ago

Two dice are rolled in a game. I win if the sum of the outcomes of the two dice is 7 or 11 and lose if the sum is 2, 3 or 12. I keep rolling until one of these sums occurs. Using conditional probabilities, what is the probability of winning this game?

There are 4 ways of getting a losing roll and 8 of getting a winning one.
The rest of the time (24/36) you roll again.
There's a (24/36)*(8/36) chance that you win on your second roll.
There's a (24/36)*(24/36)*(8/36) that you win on your third roll.
And so on.

You could some over infinity. But there's an easier way. Note that in every single round the probability of winning is exactly twice as much as the probability of losing.
Eventually the game has to end, and at every point you are twice as likely to win than to lose. Hence the chance of winning is 2/3 and losing is 1/3.

oh yeah! the game starts over if we don't roll a 2,3,7,11 or 12. And is totally memoryless of the previous rolls and that's why ending a game is just considering the win and losing ways! ohh... thanks!

If you absolutelutely need to work conditional probability into it you could take
\[S_n=\sum_{0}^{\infty}(24/36)^{n}*(8/36)\]
The n-th term in the series represents the probability that you win on exactly the n-th roll.
If I didn't mess up the value of the sum should be 2/3.